Introduction

In this blog post, we will explore how to solve the LeetCode problem "1959" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

Problem Statement

You are currently designing a dynamic array. You are given a 0-indexed integer array nums, where nums[i] is the number of elements that will be in the array at time i. In addition, you are given an integer k, the maximum number of times you can resize the array (to any size). The size of the array at time t, sizet, must be at least nums[t] because there needs to be enough space in the array to hold all the elements. The space wasted at time t is defined as sizet - nums[t], and the total space wasted is the sum of the space wasted across every time t where 0 <= t < nums.length. Return the minimum total space wasted if you can resize the array at most k times. Note: The array can have any size at the start and does not count towards the number of resizing operations. Example 1: Input: nums = [10,20], k = 0 Output: 10 Explanation: size = [20,20]. We can set the initial size to be 20. The total wasted space is (20 - 10) + (20 - 20) = 10. Example 2: Input: nums = [10,20,30], k = 1 Output: 10 Explanation: size = [20,20,30]. We can set the initial size to be 20 and resize to 30 at time 2. The total wasted space is (20 - 10) + (20 - 20) + (30 - 30) = 10. Example 3: Input: nums = [10,20,15,30,20], k = 2 Output: 15 Explanation: size = [10,20,20,30,30]. We can set the initial size to 10, resize to 20 at time 1, and resize to 30 at time 3. The total wasted space is (10 - 10) + (20 - 20) + (20 - 15) + (30 - 30) + (30 - 20) = 15. Constraints: 1 <= nums.length <= 200 1 <= nums[i] <= 106 0 <= k <= nums.length - 1

Explanation

Here's the solution to the dynamic array resizing problem:

• High-Level Approach:

  • Use dynamic programming to store the minimum wasted space for each prefix of nums given a certain number of resizes.
  • The DP state dp[i][j] represents the minimum wasted space for nums[:i] with j resizes.
  • The transition considers all possible last resize positions and calculates the wasted space incurred from that resize.

• Complexity:

  • Runtime: O(n²k), where n is the length of nums and k is the maximum number of resizes.
  • Storage: O(nk)

Code

    def minSpaceWastedKResizes(nums, k):
    n = len(nums)
    dp = [[float('inf')] * (k + 1) for _ in range(n + 1)]
    dp[0][0] = 0

    for i in range(1, n + 1):
        for j in range(k + 1):
            for p in range(i):
                max_val = 0
                total_waste = 0
                for l in range(p, i):
                    max_val = max(max_val, nums[l])
                for l in range(p, i):
                    total_waste += max_val - nums[l]
                if j == 0 and p > 0:
                    continue
                if p == 0 and j > 0:
                    continue
                if p == 0:
                    dp[i][j] = total_waste
                elif j > 0:
                     dp[i][j] = min(dp[i][j], dp[p][j - 1] + total_waste)

    return min(dp[n])